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← JEE Main 2022 · 26 Jul · Shift 1
MathematicsInteger+4 / -1Question 26 of 90

JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 26

If n(2n+1)∫01(1−xn)2ndx=1177∫01(1−xn)2n+1dxn(2n+1)\displaystyle\int_0^1\left(1-x^n\right)^{2n}dx=1177\int_0^1\left(1-x^n\right)^{2n+1}dx, then n∈Nn\in\mathbf N is equal to ______.

Answer and worked solution

Official answer: 24

Reduction. Im=mnmn+1Im−1I_{m}=\dfrac{mn}{mn+1}I_{m-1} for Im=∫01(1−xn)mdxI_m=\int_0^1(1-x^n)^mdx, so I2n+1=n(2n+1)n(2n+1)+1I2nI_{2n+1}=\dfrac{n(2n+1)}{n(2n+1)+1}I_{2n}.

Equation n(2n+1)+1=1177n(2n+1)+1=1177: n=24n=24.

Answer: 24